show that $K=\Bbb F_{q^m}$, where m is the order of $q$ in the group of units $\Bbb {Z}_n^*$ of the ring $\Bbb Z_n$.

Question

Let $q$ be power of a prime $p$, and let $n$ be a positive integer not divisible by $p$. We let $\Bbb F_q$ be the unique upto isomorphism finite field of $q$ elements. If $K$ is the splitting field of $x^n-1$ over $\Bbb F_q$, show that $K=\Bbb F_{q^m}$, where m is the order of $q$ in the group of units $\Bbb {Z_n}^*$ of the ring $\Bbb Z_n$.

Answer 1

Denote by $\mu_n \subseteq K^{\times}$ the roots of $x^n-1$. As a subgroup of the multiplicative group of a field, it is cyclic, generated by some $\zeta$. Now, $\operatorname{Gal}(K|\mathbb F_q) \hookrightarrow (\mathbb Z/n)^{\times},\ \sigma \mapsto a_{\sigma}$, where $\sigma\zeta = \zeta^{a_{\sigma}}$. But the cyclic group $\operatorname{Gal}(K|\mathbb F_q)$ is generated by the Frobenius...